I have watched Felix Baumgartner freefall; but I wonder how Felix has reached the speed of sound quickly, in a matter of some seconds, then we had no idea of its speed?
Any explanation please.
Answer
The Red Bull Stratos project involving the 43-year-old Austrian man Felix Baumgartner is to break the sound barrier. Within the first 15,000 feet of his jump he was traveling well over the cruising speed of a commercial jetliner, reaching some 625 mph. The maximum velocity reached by Felix is about some 380 km/s.
How did he do that? During a free-fall, there would be two forces acting on the object. The Air-resistance (or Drag) and Gravity ($mg$). So, Two parts come into our play...
The acceleration due to gravity $(g=\frac{GM}{(R+h)^2})$ value is almost a constant 9.8. Yes, It varies from 9.6 to 9.8 within that 39 km range. For example, at a height of 39000 m, it is some 9.684 and at a height of 10 km, it's about 9.7. At last at the sea level, it is 9.8 as you know. But, this $g$ value has no higher difference than 0.2 (even at 39 kms).
Now, the major part... The air resistance acting on a free-falling object depends on its velocity. As the velocity increases, the drag also increases and a period of time arrives when the body falls with a constant average velocity called terminal-velocity. It's probably around 50 m/s in air. At the terminal velocity, the force due to gravity equals air resistance.
But, His free-fall is from about an altitude of some 120,000 ft. (39 km) from ground. It's in the stratosphere (8-50 kms). In the stratosphere, the pressure is too low which provides the fact that the density of air is too small. This is because the weight acting upon molecules is low for higher molecules and vice-versa. This is where Drag gets weaker. 'Cause the air-resistance also depends upon the density of the medium (air). Hence, only gravity accelerates him down and he'd reach the sound barrier soon. Once he enters our troposphere (up to 8 km from ground), he slows down due to drag and he'd reach 170 m/h which is enough for parachuting, touch-down and landing at last..! On comparison to the effect caused by $g$, I'd say that air resistance plays a major role.
In addition to these, the speed of sound is also a low value at the stratosphere (a supportive factor). Because the pressure wave also depends on the density of the medium. The Laplace correction is given by $v=\sqrt{\frac{\gamma P}{\rho}}$. As the speed is inversely proportional to density $\rho$ or directly proportional to pressure, the $v$ value increases at stratosphere. (i.e.) It's not a constant 330-343 m/s. At about 30 km, it's 305 m/s. This further reduces the effect caused by $g$ difference case..!
Edit: It seems that my approximation of $g$ and $v$ has become such an issue. So, I've updated to a newer version. Some calculators I've used - Variation of $g$ with altitude and Variation of Speed of sound. If we use the Mach value 1.2, we could get the output in that calculator.
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